Question 788114
I see two ways to solve {{{h^2+28h-3=0}}}.
 
COMPLETING THE SQUARE:
{{{h^2+28h-3=0}}}
{{{h^2+28h=3}}}
Here we notice that {{{h^2+28h}}} is part of {{{h^2+28h+196=h^2+28h+14^2=(h+14)^2}}}
{{{h^2+28h+196=3+196}}}
{{{(h+14)^2=199}}}
So {{{h+14=sqrt(199)}}} or {{{h+14=sqrt(199)}}},
which can be summarized as
{{{h=-14 +- sqrt(199)}}}
 
APPLYING THE QUADRATIC FORMULA:
An equation of the form {{{ax^2+bx+c=0}}} has solutions given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
as long as {{{b^2-4*a*c>=0}}} (because otherwise {{{sqrt( b^2-4*a*c)}}} does not exist).
In the case of {{{h^2+28h-3=0}}}
{{{x=h}}}, {{{a=1}}}, {{{b=28}}}, and {{{c=-3}}}, so
{{{x = (-28 +- sqrt( 28^2-4*1*(-3) ))/(2*1) }}}
{{{x = (-28 +- sqrt(784+12 ))/2 }}}
{{{x = (-28 +- sqrt(796))/2 }}}
{{{x = (-28 +- sqrt(4*199))/2 }}}
{{{x = (-28 +- sqrt(4)*sqrt(199))/2 }}}
{{{x = (-28 +- 2sqrt(199))/2 }}}
{{{x = -14 +- sqrt(199) }}}